Linear transformations — definition, kernel, image: Worked examples

4.5.27 · D3 · Maths › Linear Algebra (Full) › Linear transformations — definition, kernel, image

Definition Naam jo hum har line pe reuse karte hain (pehle inhe build karo)

Ek linear map ko $T : V \to W$ likha jaata hai. Yahan:

$V$ = input space (ya "domain") — woh jagah jahan se hum vectors daalne waale hain.
$W$ = output space (ya "codomain") — jahan results land hone allowed hain.
Hum input space ki dimension ko $n = dim V$ bol te hain (input ke kitne independent directions hain) aur output space ki dimension ko $m = dim W$ . $R^{2} \to R^{3}$ ke liye $n = 2$ , $m = 3$ .
kernel $ker T = {v \in V : T (v) = 0}$ — woh inputs jinhein machine zero par crush kar deti hai. Yeh input space $V$ mein rehta hai.
image $im T = {T (v) : v \in V}$ — har woh output jo machine produce kar sakti hai. Yeh output space $W$ mein rehta hai.
nullity $= dim (ker T)$ , rank $= dim (im T)$ . Rank–Nullity Theorem inhe joḍta hai: $dim V = nullity + rank$ , yaani $n = nullity + rank$ .

The scenario matrix

Har linear-map problem inn cells mein se kisi ek mein aata hai. Is table ko ek checklist ki tarah padho — har row ek shape of behaviour hai. Yaad rakho $n = dim V$ (input dimension) aur $m = dim W$ (output dimension), jo upar define ho chuke hain.

#	Case class	Kya special hai	Nullity / Rank	Example
A	Injective, not onto ( $n < m$ : input dims output se kam)	dimension mein upar jaata hai, kuch crush nahi karta	null $= 0$ , rank $= n$	Ex 1
B	Onto, not injective ( $n > m$ : input dims output se zyada)	neeche map karta hai, kuch crush karna hi padega	null $> 0$ , rank $= m$	Ex 2
C	Bijective (square, $n = m$ , invertible)	perfect one-to-one, dono directions	null $= 0$ , rank $= n$	Ex 3
D	The zero map (degenerate)	sab kuch crush kar deta hai	null $= n$ , rank $= 0$	Ex 4
E	Projection (idempotent, kernel = ek line/plane)	ek subspace ko rakhta hai, baaki ko khatam karta hai	null $> 0$ , rank $< n$	Ex 5
F	Rank-deficient square ( $n = m$ par $det = 0$ )	square par invertible NAHI	null $> 0$ , rank $< n$	Ex 6
G	Non- $R^{n}$ space (polynomials, derivative)	vectors columns nahi hain	null $= 1$ , rank $= n - 1$	Ex 7
H	Word problem (real world)	pehle translate karo, phir compute karo	depends	Ex 8
I	Exam twist (find $k$ so a vector is reachable)	image-membership test	—	Ex 9

Injective surjective bijective properties bas cells A, B, C hain jo nullity/rank se seedha padh li jaati hain.

Example 1 — Cell A: injective, not onto ( $n = 2 < m = 3$ )

Kernel: output ko zero set karo: $x = 0$ , $y = 0$ , $x + y = 0$ . Pehle do hi $x = y = 0$ force kar dete hain. Yeh step kyun? Kernel = woh inputs jo $0$ par bheje jaate hain; seedha $T (x, y) = (0, 0, 0)$ solve karo.
Toh $ker T = {(0, 0)}$ , nullity $= 0$ → $T$ injective hai. Yeh step kyun? Injective $⟺ ker T = {0}$ (parent ka rule, koi exception nahi).
Image: matrix hai $A = 101011$ ; columns $(1, 0, 1)$ aur $(0, 1, 1)$ . Yeh ek doosre ke multiples nahi hain, toh independent hain → rank $= 2$ . Yeh step kyun? $T (x) = A x$ ke liye, image column space hai; uski dimension independent columns ki sankhya hai.
Kyunki rank $= 2 < 3 = dim R^{3}$ , image 3D ke andar ek plane hai, poori nahi → not onto.

Figure — Linear transformations — definition, kernel, image

Example 2 — Cell B: onto, not injective ( $n = 3 > m = 2$ )

Kernel: $x - z = 0$ aur $y + z = 0$ , toh $x = z$ aur $y = - z$ . $z = t$ maano: $(t, - t, t) = t (1, - 1, 1)$ . Yeh step kyun? $T = 0$ solve karo; ek free parameter $t$ bachi rehti hai.
$ker T = span {(1, - 1, 1)}$ , nullity $= 1$ . Nonzero kernel → not injective. Yeh step kyun? Ek free variable = inputs ki poori ek line $0$ par crush hoti hai.
Image: matrix $A = (1001 - 1 1)$ ; columns $(1, 0), (0, 1), (- 1, 1)$ . Pehle do pehle se hi poore $R^{2}$ ko span karte hain → rank $= 2$ , image $= R^{2}$ → onto. Yeh step kyun? Do independent columns 2D output space fill kar dete hain.

Example 3 — Cell C: bijective (invertible, $n = m = 2$ )

Matrix $A = (2111)$ , determinant $= 2 \cdot 1 - 1 \cdot 1 = 1 \neq = 0$ . Yeh step kyun? Nonzero determinant matlab columns independent hain — koi collapse nahi.
Kernel: $2 x + y = 0$ , $x + y = 0$ . Subtract karo: $x = 0$ , phir $y = 0$ . Toh $ker T = {0}$ , nullity $= 0$ , injective. Yeh step kyun? $T = 0$ solve karo; unique solution ka $0$ hona $det \neq = 0$ confirm karta hai.
Image: rank $= 2$ (do independent columns) $= dim R^{2}$ → onto. Injective + onto = bijective.

Example 4 — Cell D: the zero map (degenerate)

Kernel: har input $0$ deta hai, toh $ker T = R^{2}$ , nullity $= 2$ . Yeh step kyun? Kernel woh set hai jo $0$ par bheja jaata hai — yahan woh poori space hai.
Image: sirf output $(0, 0)$ hai, toh $im T = {0}$ , rank $= 0$ . Yeh step kyun? Kuch aur nahi aa sakta.
Injective nahi (kernel $\neq = {0}$ ), onto nahi (image $\neq = R^{2}$ ).

Example 5 — Cell E: projection (idempotent)

Kernel: $T (x, y) = (0, 0)$ ke liye $x = 0$ chahiye; $y$ free hai. Toh $ker T = {(0, y)} = span {(0, 1)}$ , yaani $y$ -axis, nullity $= 1$ . Yeh step kyun? Origin ke seedha upar/neeche ke points apna shadow $0$ par daalte hain.
Image: outputs $(x, 0)$ hain = poori $x$ -axis, rank $= 1$ . Yeh step kyun? Har shadow $x$ -axis par hi padta hai; uske upar kuch reachable nahi.
Idempotent check: $T (T (x, y)) = T (x, 0) = (x, 0) = T (x, y)$ . Do baar project karna = ek baar project karna. ✓ Yeh step kyun? Ek projection jo already uski image mein hai use fix kar deta hai (yahi upar idempotent ki definition hai).

Example 6 — Cell F: rank-deficient square ( $n = m = 2$ , $det = 0$ )

$det A = 1 \cdot 4 - 2 \cdot 2 = 0$ . Invertible nahi. Yeh step kyun? Zero determinant dependent columns flag karta hai → ek collapse.
Kernel: $x + 2 y = 0$ (dono rows yahi deti hain), toh $x = - 2 y$ . $(- 2 y, y) = y (- 2, 1)$ . $ker T = span {(- 2, 1)}$ , nullity $= 1$ . Yeh step kyun? Dono equations secretly ek hi equation hain → ek free variable.
Image: columns $(1, 2)$ aur $(2, 4)$ ; doosra pehle ka $2 \times$ hai → sirf ek independent column. $im T = span {(1, 2)}$ , rank $= 1$ — ek line, plane nahi. Yeh step kyun? Dependent columns lower-dimensional space span karte hain.

Example 7 — Cell G: polynomials par derivative

Kernel: $D (a + b x + c x^{2}) = b + 2 c x = 0$ ke liye $b = 0$ aur $c = 0$ zaroori hai; $a$ free hai. Toh $ker D = {a \cdot 1} =$ constants, nullity $= 1$ . Yeh step kyun? Sirf constant functions ka slope har jagah zero hota hai.
Image: $D (a + b x + c x^{2}) = b + 2 c x$ , jo degree $\leq 1$ ka koi bhi polynomial hai = $span {1, x}$ , rank $= 2$ . Yeh step kyun? Differentiate karne se degree ek kam ho jaati hai, toh outputs $P_{1}$ mein rehte hain.

Example 8 — Cell H: word problem

Likho $T (a, b, c) = (a + b + c, 2 a + b + 3 c)$ , matrix $A = (121113)$ . Yeh step kyun? Dono recipe formulas se seedhe coefficients padho.
(i) Kernel: $a + b + c = 0$ aur $2 a + b + 3 c = 0$ . Pehle se doosra subtract karo: $a + 2 c = 0 \Rightarrow a = - 2 c$ . Phir $b = - a - c = 2 c - c = c$ . Toh $(a, b, c) = (- 2 c, c, c) = c (- 2, 1, 1)$ . Yeh step kyun? Ek variable eliminate karo taaki free parameter samne aaye.
$ker T = span {(- 2, 1, 1)}$ : is direction ke blends algebraically zero volume aur zero sugar cancel karte hain. Nullity $= 1$ .
(ii) Image: columns $(1, 2)$ aur $(1, 1)$ independent hain → rank $= 2$ = $dim R^{2}$ . Har (volume, sugar) pair reachable hai → onto.

Example 9 — Cell I: exam twist (image membership)

Hume kuch $x, y$ ke liye $(x + y, x - y, 2 x) = (3, 1, k)$ chahiye. Pehle do se: $x + y = 3$ , $x - y = 1$ → joḍo: $2 x = 4$ , $x = 2$ , phir $y = 1$ . Yeh step kyun? Pehle "free" coordinates solve karo; woh $x, y$ ko uniquely pin kar dete hain.
Teesra coordinate forced hai: $2 x = 2 \cdot 2 = 4$ . Toh hume $k = 4$ chahiye. Yeh step kyun? Ek baar $x$ fix ho jaaye, teesra output independently choose nahi ho sakta — plane ki constraint.
Isliye $(3, 1, k) \in im T ⟺ k = 4$ .

Active recall

Recall Har map kaun sa cell hai? (answers hide karo)

$R^{4} \to R^{2}$ , onto ::: Cell B — onto, not injective (nullity $= 2$ ). $T (v) = 0$ ::: Cell D — the zero map (nullity = dim $V$ , rank $0$ ). $2 \times 2$ with $det = 0$ ::: Cell F — rank-deficient square. Derivative on $P_{3}$ ::: Cell G — non- $R^{n}$ ; nullity $1$ , rank $3$ . $R^{2} \to R^{5}$ injective ::: Cell A — injective, not onto. Idempotent ka matlab kya hai? ::: $T \circ T = T$ — do baar apply karna = ek baar apply karna (projections).

Connections

Column space and null space — yahan har "image" ek column space hai, har "kernel" ek null space.
Rank–Nullity Theorem — saare nine examples ka running verify.
Injective surjective bijective — cells A/B/C yahi teen properties hain jo nullity aur rank se padhi jaati hain.
Eigenvalues and eigenvectors — zero determinant (Cell F) matlab $0$ ek eigenvalue hai; uske eigenvectors kernel banate hain.
Change of basis — abstract derivative map (Ex 7) ko concrete matrix mein badalta hai.
Matrix multiplication — do linear maps compose karna unke matrices multiply karta hai.